← 2019 Paper 2

UPSC 2019 Maths Optional Paper 2 Q1e — Step-by-Step Solution

10 marks · Section A

Graphical method · Linear Programming · asked 5× in 14 yrs · Read the full method →

Question

Use graphical method to solve the linear programming problem. Maximize Z=3x1+2x2Z=3x_1+2x_2 subject to

x1−x2≥1,x1+x3≥3\begin{aligned}x_1-x_2&\ge 1,\\ x_1+x_3&\ge 3\end{aligned}

and x1,x2,x3≥0x_1,x_2,x_3\ge 0.

Technique

Graphical method; recognize the redundant slack-like variable x3x_3, reduce to the plane, identify the single vertex and an unbounded edge along which Z→∞Z\to\infty.

Solution

Step 1 — Reduce to a planar (graphical) problem

The objective Z=3x1+2x2Z=3x_1+2x_2 does not involve x3x_3. The variable x3x_3 appears only in the second constraint x1+x3≥3x_1+x_3\ge 3. Since x3≥0x_3\ge0 is free to be as large as we like, the constraint x1+x3≥3x_1+x_3\ge3 can always be satisfied for any x1≥0x_1\ge0 (choose x3≥3−x1x_3\ge 3-x_1, e.g. x3=max⁡(0,3−x1)x_3=\max(0,3-x_1)). Hence the second constraint imposes no restriction on the objective variables (x1,x2)(x_1,x_2).

The problem therefore reduces, in the (x1,x2)(x_1,x_2)-plane, to:

maximize Z=3x1+2x2s.t.x1−x2≥1,  x1≥0, x2≥0.\text{maximize } Z=3x_1+2x_2 \quad\text{s.t.}\quad x_1-x_2\ge1,\ \ x_1\ge0,\ x_2\ge0.

Step 2 — Feasible region in the (x1,x2)(x_1,x_2)-plane

x1−x2≥1x_1-x_2\ge1 is the half-plane on/below the line x2=x1−1x_2=x_1-1 (intercept (1,0)(1,0), slope 11), intersected with the first quadrant. The boundary line meets the x1x_1-axis at (1,0)(1,0). The region is the set

{(x1,x2):x1≥0, x2≥0, x2≤x1−1},\{(x_1,x_2):x_1\ge0,\ x_2\ge0,\ x_2\le x_1-1\},

a wedge opening to the right, unbounded in the +x1+x_1 (and along it +x2+x_2) direction. Its only vertex is (1,0)(1,0).

Step 3 — Behaviour of the objective on the region

Move along the edge x2=x1−1x_2=x_1-1 (a feasible boundary, with x2≥0⇔x1≥1x_2\ge0\Leftrightarrow x_1\ge1):

Z=3x1+2(x1−1)=5x1−2→+∞as x1→∞.Z=3x_1+2(x_1-1)=5x_1-2\to+\infty\quad\text{as }x_1\to\infty.

So ZZ can be made arbitrarily large within the feasible region.

Answer

  The LPP has an unbounded solution: max⁡Z=+∞ (no finite optimum).  \boxed{\;\text{The LPP has an }\textbf{unbounded solution}\text{: }\max Z=+\infty\ (\text{no finite optimum}).\;}
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